Probability Ques 117
Passage Based Problems
Let $n _1$ and $n _2$ be the number of red and black balls, respectively in box I. Let $n _3$ and $n _4$ be the number of red and black balls, respectively in box II.
(2015 Adv.)
One of the two boxes, box I and box II was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II, is $\frac{1}{3}$, then the correct option(s) with the possible values of $n _1, n _2, n _3$ and $n _4$ is/are
(a) $n _1=3, n _2=3, n _3=5, n _4=15$
(b) $n _1=3, n _2=6, n _3=10, n _4=50$
(c) $n _1=8, n _2=6, n _3=5, n _4=20$
(d) $n _1=6, n _2=12, n _3=5, n _4=20$
Show Answer
Answer:
Correct Answer: 117.(b)
Solution:
$ \begin{array}{|ll|} n _1 & \text { Red } \\ n _2 & \text { Black } \end{array} \underbrace{\left|\begin{array}{ll} n _3 & \text { Red } \\ n _4 & \text { Black } \end{array}\right|} $
Let $\quad A=$ Drawing red ball
$ \therefore \quad P(A)=P\left(B _1\right) \cdot P\left(A / B _1\right)+P\left(B _2\right) \cdot P\left(A / B _2\right) $
$ =\frac{1}{2} (\frac{n _1}{n _1+n _2})+\frac{1}{2} (\frac{n _3}{n _3+n _4}) $
Given, $\quad P\left(B _2 / A\right)=\frac{1}{3}$
$ \begin{aligned} & \Rightarrow \quad \frac{P\left(B _2\right) \cdot P\left(B _2 \cap A\right)}{P(A)}=\frac{1}{3} \\ & \Rightarrow \quad \frac{\frac{1}{2} (\frac{n _3}{n _3+n _4})}{\frac{1}{2} (\frac{n _1}{n _1+n _2})+\frac{1}{2} (\frac{n _3}{n _3+n _4})}=\frac{1}{3} \\ & \Rightarrow \quad \frac{n _3\left(n _1+n _2\right)}{n _1\left(n _3+n _4\right)+n _3\left(n _1+n _2\right)}=\frac{1}{3} \end{aligned} $
Now, check options, then clearly options (a) and (b) satisfy.
BETA
